Question

Arrange $\sqrt[3]{327},\sqrt[3]{3343},4\sqrt{5},\sqrt{96}$ in ascending order.

• A. $\sqrt[3]{327}<\sqrt[3]{3343}<4\sqrt{5}<\sqrt{96}$
• B. $\sqrt[3]{327}>\sqrt[3]{3343}>4\sqrt{5}>\sqrt{96}$
• C. $\sqrt[3]{3343}>\sqrt{96}>\sqrt[3]{327}>4\sqrt{5}$
• D. $\sqrt[3]{3343}<\sqrt{96}<4\sqrt[3]{35}<\sqrt{27}$

Answer 1

The correct option is A $\sqrt[3]{327}<\sqrt[3]{3343}<4\sqrt{5}<\sqrt{96}$

First we convert the numbers into their perfect cube and square respectively.
So we get,
$\sqrt[3]{327}=\sqrt[3]{33\times 3\times 3}=3\&\sqrt[3]{3343}=\sqrt[3]{37\times 7\times 7}=7$

$\therefore \sqrt[3]{3343}>\sqrt[3]{327}$

Now, approximating value of $4\sqrt{5}$
$\because 4.84<5.00<5.29\Rightarrow (2.2{)}^2<5.00<(2.3{)}^2\Rightarrow 2.2<\sqrt{5}<2.3\Rightarrow 8.8<4\sqrt{5}<9.2$

Also, approximating $\sqrt{96}=\sqrt{16\times 6}=4\sqrt{6}$
$\because 5.76<6.00<6.25\Rightarrow (2.4{)}^2<6.00<(2.5{)}^2\Rightarrow 2.4<\sqrt{6}<2.5\Rightarrow 9.6<4\sqrt{6}<10.0$

$\Rightarrow 4\sqrt{6}>9.6\&4\sqrt{5}<9.2$

Clearly it can be seen that $3<7<4\sqrt{5}<4\sqrt{6}$
$\therefore \sqrt[3]{327}<\sqrt[3]{3343}<4\sqrt{5}<\sqrt{96}$